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# Algebra

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Let y = (x-3)/(x+7) + 1/3

Then this equation can be expressed in the form (x + a)(y + b) = c for some constants a, b, and c . Enter your answer in the form " a, b, c".

Jul 11, 2022

### 1+0 Answers

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First, note that $$(x+a)(y+b) = xy + xb + ay + ab$$

Start by simplifying the RHS:

$$={x - 3 \over x + 7} + {1 \over 3}$$

$$= { 3(x - 3) \over 3(x + 7) } + {x + 7 \over 3x + 21}$$

$$= { 3x-9 \over 3x+21 } + {x + 7 \over 3x + 21}$$

$$= { 4x - 2\over 3x + 21}$$

Now, we have the equation $$y = {4x - 2\over 3x + 21}$$

Then, cross multiply to get $$3xy + 21y = 4x - 2$$

Subtract 4x from both sides: $$3xy + 21y - 4x =- 2$$

Divide by 3: $$xy + 7y - {4 \over 3}x = -{2 \over 3}$$

Now, recall the factored form. We know that $$ay = 7y$$ and $$bx = -{4 \over 3}x$$, meaning $$(a,b) = \ (7, -{4 \over 3})$$

But, remember that we need to add $$a \times b = 7 \times -{4 \over 3} = -{28 \over 3}$$, so we can factor. Thus, $$c = -{2 \over 3} -{28 \over 3} = -10$$

This means it is $$\color{brown}\boxed{(x + 7) (y - {4 \over 3})=-10}$$

Jul 11, 2022